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Let $T\colon X\to X$ be an upper-semi Fredholm operator acting on a $B$-space $X$ (the range of $T$ is closed and kernel is finite-dimensional) with complemented range. Suppose $S\colon X\to X$ is bounded below. Does it follow that $T+S$ has complemented range? If we have assumed that $S$ is compact, then the answer would be 'yes'.

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What do you mean by bounded below? –  Kevin Beanland Mar 28 '12 at 16:36
    
Presumably on the unit sphere, Kevin; hence an isomorphism into. –  Bill Johnson Mar 28 '12 at 18:31
    
That's what I thought, but I got confused since he considered compact S in the next sentence. –  Kevin Beanland Mar 28 '12 at 19:59

2 Answers 2

up vote 3 down vote accepted

In fact, any bounded operator $L$ on $X$ may be written as a sum of two invertible operators $S$ and $T$, hence in particular both Fredholm and bounded below. We may take $S:=\lambda I$ and $T:=L-\lambda I$, with $\lambda > \|L\|$.

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This answer is a good reason that I should post anonymously. :) –  Bill Johnson Mar 28 '12 at 20:21
    
On the contrary, since you are a source of knowledge for all of us, you are well above whatever we post here :) –  Pietro Majer Mar 29 '12 at 4:32

No. Take $U$ mapping $X$ isomorphically onto a subspace which is complemented via a projection $P$ and $V$ mapping $X$ isomorphically onto an uncomplemented subspace s.t. $PV=0$. (This situation is easily realizable in many spaces; e.g., having $X$ isomorphic to $X\oplus X$ and having $X$ contain an uncomplemented subspace isomorphic to itself is sufficient. Take for $U$ an isomorphism from $X\oplus X$ onto $X\oplus 0$ and let $V$ be an isomorphism from $X\oplus X$ onto an uncomplemented subspace of $0 \oplus X$ and let $P$ be the natural projection form $X\oplus X$ onto $X\oplus 0$.) Set $T=U+V$ and $S=-U$.

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